Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Got this one right however was wondering if anybody knows how to proceed the algabraic way.. Maybe we can do something like this; x+ x+1 + x+2 + x+3=5K 4x+6=5K ..?

Re: sum of 4 consecutive positive integers [#permalink]

Show Tags

28 Sep 2009, 21:49

1

This post received KUDOS

4X+6=5k

1, x is odd, no hint, wrong

2, none of x-> x+3 is divisible by 5, If not

x is divisible by 5 then 4x+6 divides 5 remains 1 not 0 x+1 is ............5 then sum=4(x+1)+2 divides 5 remains 2 not 0 x+2 is ............5 then sum=4(x+2)-2 divides 5 remains 3 not 0 x+3 is.............5 then sum=4(x+3)-6 divides 5 remains 4 not 0

2 is right

3, 1 Number with the digit is 3, the other should be ended with + 0,1,2 then sum =10k+6 not divisible by 5 + 1,2,4 then sum =10K+10 divisible by 5 these above enough to answer + 2,4,5 then + 4,5,6

Re: sum of 4 consecutive positive integers [#permalink]

Show Tags

29 Sep 2009, 04:14

1

This post received KUDOS

tejal777 wrote:

If the sum of 4 consecutive positive integers is divisible by 5, which of the following must be true?

I. The smallest integer is odd

II. None of the integers is divisible by 5

III. One of the integers has 3 as its units digit

Got this one right however was wondering if anybody knows how to proceed the algabraic way..

I typically think in terms of remainders. If x = 5*k+y (where y = 0,1,2,3 or 4; sometimes I use y = -2,-1,0,1,2). In the above problem x can be replaced with y. it's pretty obvious that the neither of the 4 numbers can be divisible by 5. Otherwise, 4 options -2,-1,0,1 or -1,0,1,2 or 0,1,2,3 (-2) or -3(2), -2,-1,0. clearly, neither of those options work. necessary and sufficient condition is: neither of the 4 numbers is divisible by 5.

I. The smallest integer is odd -> false (6,7,8,9)

II. None of the integers is divisible by 5 -> true (from above)

III. One of the integers has 3 as its units digit -> false (6,7,8,9)